Av(1342, 2413, 3142)
Counting Sequence
1, 1, 2, 6, 21, 79, 310, 1251, 5151, 21536, 91137, 389510, 1678565, 7284975, 31811311, ...
Implicit Equation for the Generating Function
\(\displaystyle x F \left(x
\right)^{4}+\left(-x -1\right) F \left(x
\right)^{3}+\left(x^{2}-2 x +3\right) F \left(x
\right)^{2}+\left(2 x -3\right) F \! \left(x \right)+1 = 0\)
Recurrence
\(\displaystyle a \! \left(0\right) = 1\)
\(\displaystyle a \! \left(1\right) = 1\)
\(\displaystyle a \! \left(2\right) = 2\)
\(\displaystyle a \! \left(3\right) = 6\)
\(\displaystyle a \! \left(4\right) = 21\)
\(\displaystyle a \! \left(5\right) = 79\)
\(\displaystyle a \! \left(6\right) = 310\)
\(\displaystyle a \! \left(7\right) = 1251\)
\(\displaystyle a \! \left(8\right) = 5151\)
\(\displaystyle a \! \left(n +9\right) = \frac{n \left(2 n +3\right) \left(2 n -1\right) a \! \left(n \right)}{\left(n +10\right) \left(n +8\right) \left(2 n +17\right)}+\frac{\left(8 n^{3}+83 n^{2}+145 n +55\right) a \! \left(n +1\right)}{\left(n +10\right) \left(n +8\right) \left(2 n +17\right)}+\frac{\left(1151 n^{3}+8520 n^{2}+22579 n +18546\right) a \! \left(n +2\right)}{12 \left(n +10\right) \left(n +8\right) \left(2 n +17\right)}-\frac{\left(2233 n^{3}+55122 n^{2}+279635 n +395634\right) a \! \left(n +3\right)}{24 \left(n +10\right) \left(n +8\right) \left(2 n +17\right)}+\frac{\left(62879 n^{3}+790140 n^{2}+3299443 n +4570050\right) a \! \left(n +4\right)}{24 \left(n +10\right) \left(n +8\right) \left(2 n +17\right)}-\frac{\left(67265 n^{3}+1013004 n^{2}+5056489 n +8372154\right) a \! \left(n +5\right)}{24 \left(n +10\right) \left(n +8\right) \left(2 n +17\right)}+\frac{\left(7612 n^{3}+136431 n^{2}+810569 n +1597989\right) a \! \left(n +6\right)}{6 \left(n +10\right) \left(n +8\right) \left(2 n +17\right)}-\frac{\left(3640 n^{3}+76029 n^{2}+526883 n +1212402\right) a \! \left(n +7\right)}{12 \left(n +10\right) \left(n +8\right) \left(2 n +17\right)}+\frac{\left(458 n^{3}+10893 n^{2}+85993 n +225450\right) a \! \left(n +8\right)}{12 \left(n +10\right) \left(n +8\right) \left(2 n +17\right)}, \quad n \geq 9\)
\(\displaystyle a \! \left(1\right) = 1\)
\(\displaystyle a \! \left(2\right) = 2\)
\(\displaystyle a \! \left(3\right) = 6\)
\(\displaystyle a \! \left(4\right) = 21\)
\(\displaystyle a \! \left(5\right) = 79\)
\(\displaystyle a \! \left(6\right) = 310\)
\(\displaystyle a \! \left(7\right) = 1251\)
\(\displaystyle a \! \left(8\right) = 5151\)
\(\displaystyle a \! \left(n +9\right) = \frac{n \left(2 n +3\right) \left(2 n -1\right) a \! \left(n \right)}{\left(n +10\right) \left(n +8\right) \left(2 n +17\right)}+\frac{\left(8 n^{3}+83 n^{2}+145 n +55\right) a \! \left(n +1\right)}{\left(n +10\right) \left(n +8\right) \left(2 n +17\right)}+\frac{\left(1151 n^{3}+8520 n^{2}+22579 n +18546\right) a \! \left(n +2\right)}{12 \left(n +10\right) \left(n +8\right) \left(2 n +17\right)}-\frac{\left(2233 n^{3}+55122 n^{2}+279635 n +395634\right) a \! \left(n +3\right)}{24 \left(n +10\right) \left(n +8\right) \left(2 n +17\right)}+\frac{\left(62879 n^{3}+790140 n^{2}+3299443 n +4570050\right) a \! \left(n +4\right)}{24 \left(n +10\right) \left(n +8\right) \left(2 n +17\right)}-\frac{\left(67265 n^{3}+1013004 n^{2}+5056489 n +8372154\right) a \! \left(n +5\right)}{24 \left(n +10\right) \left(n +8\right) \left(2 n +17\right)}+\frac{\left(7612 n^{3}+136431 n^{2}+810569 n +1597989\right) a \! \left(n +6\right)}{6 \left(n +10\right) \left(n +8\right) \left(2 n +17\right)}-\frac{\left(3640 n^{3}+76029 n^{2}+526883 n +1212402\right) a \! \left(n +7\right)}{12 \left(n +10\right) \left(n +8\right) \left(2 n +17\right)}+\frac{\left(458 n^{3}+10893 n^{2}+85993 n +225450\right) a \! \left(n +8\right)}{12 \left(n +10\right) \left(n +8\right) \left(2 n +17\right)}, \quad n \geq 9\)
This specification was found using the strategy pack "Point Placements" and has 11 rules.
Found on July 23, 2021.Finding the specification took 1 seconds.
Copy 11 equations to clipboard:
\(\begin{align*}
F_{0}\! \left(x \right) &= F_{1}\! \left(x \right)+F_{2}\! \left(x \right)\\
F_{1}\! \left(x \right) &= 1\\
F_{2}\! \left(x \right) &= F_{3}\! \left(x \right)\\
F_{3}\! \left(x \right) &= F_{4}\! \left(x \right) F_{7}\! \left(x \right)\\
F_{4}\! \left(x \right) &= F_{0}\! \left(x \right)+F_{5}\! \left(x \right)\\
F_{5}\! \left(x \right) &= F_{6}\! \left(x \right)\\
F_{6}\! \left(x \right) &= F_{0}\! \left(x \right) F_{4}\! \left(x \right) F_{7}\! \left(x \right) F_{8}\! \left(x \right)\\
F_{7}\! \left(x \right) &= x\\
F_{8}\! \left(x \right) &= F_{1}\! \left(x \right)+F_{9}\! \left(x \right)\\
F_{9}\! \left(x \right) &= F_{10}\! \left(x \right)\\
F_{10}\! \left(x \right) &= F_{8} \left(x \right)^{2} F_{7}\! \left(x \right)\\
\end{align*}\)